an object is moving on a horizontal frictionless surface. if the net force applied to the object in the direction of motion is doubled, the magnitude of the acceleration of the object is
1. halved
2. doubled
3. unchanged
4. quadrupled

In a tractor pull, a tractor put 250,000 J of work into pulling a large mass. The tractor pulls the mass using 98,000N of force. The tractor pulled the mass to a distance of 2.55 meters

We may use the work done formula to solve this problem:

Work = Force  x Distance x Cosine (angle between force and displacement)

Yet, because the force and displacement are applied in the same direction, the angle between them is zero, and the cosine of zero is one. As a result, we may reduce the formula to:

Work = Force x Distance

Because we know the work done is 250,000 J and the force exerted is 98,000 N, we can rewrite the formula to solve for distance:

Distance = Work / Force

Distance = 250,000 J / 98,000 N

Distance = 2.55 meters

As a result, The tractor pulled the mass to a distance of 2.55 meters

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Answer:

D

i just gave it a try i dont really know

Answer:

x-component of velocity: 7.5 m/s

y-component of velocity: 13 m/s

Explanation:

This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.

By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

Answer:

B.-80.35 J

i dont know the calculation

Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

where;

F is the applied force

v is velocity = d/t

P = F x (d/t)

P = 32 x (0.8 /0.5)

P = 32 x 1.6

P = 51.2 W

Therefore, the power exerted by the student is 51.2 W

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = f × d

Where f is force applied and d is distance travelled.

Given that;

W = f × d

3500kgm²/s² = f × 1.5m

f = 3500kgm²/s² ÷ 1.5m

f = 2.3 × 10³ kgm/s²

f = 2.3 × 10³ N

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

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uhhhhhhhhhhhhhhhhhHHhhhHhhHHhHh

The frequency of oscillation of the swing is 0.25 Hz.

The time period of oscillation of the swing is 4 s.

The length of the vine of the swing is 3.97 m.

The restoring force acting on Mohammed is 692.9 N.

Mass of Mohammed, m = 50 kg

Angle made by the vine with the vertical, θ = 45°

Number of complete swings made by Mohammed, n = 30

Time taken for this swing, t = 2 minutes = 120 seconds

a) The frequency of the swing is defined as the number of complete oscillations in one second.

So, the frequency of oscillation of the swing is,

f = n/t

f = 30/120

f = 0.25 Hz

b) The time period of oscillation of the swing is,

T = 1/f

T = 1/0.25

T = 4 s

c) The expression for the time period is given by,

T = 2π√(l/g)

T² = 4π² x (l/g)

l/g = T²/4π²

Therefore, the length of the vine of the swing is,

l = T²g/4π²

l = 4² x 9.8/4 x (3.14)²

l = 3.97 m

d) The restoring force acting on Mohammed,

F = mg sinθ

F = 50 x 9.8 x sin 45°

F = 490 x 1/√2 = 490/1.414

F = 692.9 N

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The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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Answer:

Basic kinematics, negating drag and assuming ideal conditions, we use the equation:

d=vi*t+1/2*a*t^2

Since vi is 0 (we know this because you’re dropping it, not throwing it)…

…and the only acceleration acting on it is gravity, a=9.8 m/s^2…

…we get

d=1/2(9.8)(5)^2

Explanation:

Some quick mental math tells us that this is about 125 m.

Plugging it in, we find it to be 122.5 m.

a) The final pressure and temperature for the isothermal compression are \(4.5*10^5 Pa\) and 293 K, respectively, while  b) the final pressure and temperature for the adiabatic compression are\(5.58*10^5 Pa\) and 515 K, respectively.

a. Isothermal compression:

For an isothermal process, the temperature remains constant. Therefore, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.

Since the process is isothermal, we can write:

\(P_1V_1 = P_2V_2\)

where P1 and V1 are the initial pressure and volume, and\(P_2\)and\(V_2\)are the final pressure and volume.

We are given that the volume is compressed to 1/3 of its original volume, so\(V_2 = (1/3)V_1\). Substituting this into the equation above gives:

\(P_2 = (V_1/V_2)P_1 = 3P_1\) = \(4.5*10^5 Pa\)

To find the final temperature, we can use the ideal gas law again:

PV = nRT

Rearranging, we get:

T = PV/(nR)

Substituting the values we know, we get:

T = (\(1.5*10^5\)Pa)(V1)/(nR)

Since the process is isothermal, the temperature remains constant, so the final temperature is the same as the initial temperature:

T2 = T1 = 293 K

b. Adiabatic compression:

For an adiabatic process, there is no heat transfer between the gas and its surroundings. Therefore, we can use the adiabatic equation:

PV^γ = constant

where γ = Cp/Cv is the ratio of specific heats.

Since the process is adiabatic and reversible, we can write:

\(P_1V_1\)^γ = \(P_2V_2\)^γ

We are given that the volume is compressed to 1/3 of its original volume, so V2 = (1/3)V1. Substituting this into the equation above gives:

\(P_2 = P_1(V_1/V_2)\)^γ = \(P_1\)\((3)^{(1.4)\) = \(5.58*10^5 Pa\)

To find the final temperature, we can use the adiabatic equation again:

\(T_2 = T_1(P_2/P_1)\)^((γ-1)/γ) = T1(5.58/1.5)^(0.4) = 515 K

Therefore, the final pressure and temperature for the isothermal compression are \(4.5*10^5 Pa\)and 293 K, respectively, while the final pressure and temperature for the adiabatic compression are \(5.58*10^5\) Pa and 515 K, respectively.

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Answer:

E. Hydrogen

Explanation:

It is important to note that since Hydrogen is categorized as the lightest known element in the universe today. It is therefore the only gas molecules among the listed option that would be most likely be moving at a speed high enough to escape the earth's atmosphere.

Because Hydrogen has just a single proton and a single electron this makes it easy to acquire enough energy to pull the earth's gravity.

The gas that is most likely to attain a speed greater than the earth's escape velocity is hydrogen.

The root - mean - square velocity of a gas depends on the temperature as well as the molar mass of the gas. We are told that the escape velocity of the earth is  11 km/s.

The gas that will move at such speed must be a very light gas. Hence the gas that is most likely to attain a speed greater than the earth's escape velocity is hydrogen.

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The electrical conductivity of copper at 300 K is 0.566 x 10⁸ Sm⁻¹.

Thermal conductivity and electrical conductivity are two different physical quantities used to describe the ability of a material to conduct heat and electricity respectively. Copper is a metal known to be an excellent conductor of both heat and electricity.

The relationship between thermal conductivity and electrical conductivity in copper is given by the Wiedemann-Franz Law, which states that the ratio of the electrical conductivity to thermal conductivity of a metal is proportional to its temperature.

To calculate the electrical conductivity of copper at 300 K, we will make use of the Wiedemann-Franz Law. The law states that:σ / κ = L T Where σ is the electrical conductivity, κ is the thermal conductivity, L is the Lorenz number, and T is the temperature of the material.

Substituting the values given in the problem, we get:σ / 470.4 = (2.45 x 10⁻⁸) x 300σ = (2.45 x 10⁻⁸) x 300 x 470.4σ = 0.566 x 10⁸ Sm⁻¹Therefore, the electrical conductivity of copper at 300 K is 0.566 x 10⁸ Sm⁻¹.

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Answer:

Half

Explanation:

If the half life of an isotope is 20 years, then half of the original amount will  remain after 20 years has passed.

That's what "half-life" means.

Similarly, after ANOTHER 20 years, 1\4 of the original amount remains.

And after ANOTHER 20 years, 1\8 of the original amount remains.

And after ANOTHER 20 years, 1\16 of the original amount remains.

And after ANOTHER 20 years, 1\32 of the original amount remains.

And after ANOTHER 20 years, 1\64 of the original amount remains.

And after ANOTHER 20 years, 1\128 of the original amount remains.

With an initial horizontal velocity of 32.31 m/s and a time of 3.23 seconds to impact the earth, the projectile was launched.

When an object is hurled horizontally in projectile motion, it has zero initial vertical velocity and only a horizontal initial velocity. Through the projectile's entire motion, the original horizontal velocity is maintained.

\(y = 1/2 * a_y * t^2\)

\(t^2 = 2y / a_y\)

\(t = sqrt(2y / a_y)t = sqrt(2*75.72 m / 9.81 m/s^2)\)

t = 3.23 s (rounded to two decimal places)

\(x = vi_x * t\)

\(vi_x = x / t\)

\(vi_x = 104.42 m / 3.23 s\)

\(vi_x = 32.31 m/s\)(rounded to two decimal places)

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Answer:

it is double the temperature change of iron

The top surface moves relative to the bottom surface by 1.167 x 10⁻⁶ m.

The distance the top surface move relative to the bottom surface is calculated as follows;

shear strain = F/(αx)

where;

shear strain = (14 )/(1.2 x 10⁷ x 0.00117 m)

shear strain = 0.000997

The movement of the top surface;

top surface movement = shear strain x thickness

= 0.000997  x 0.00117 m

= 1.167 x 10⁻⁶ m

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The radii of the first five Fresnel zones is 3.6 mm.

Distance from the light source to the wave surface, d₁ = 1 m

Distance from the wave surface to the observation point, d₂ = 1 m.

Wavelength of the light used, λ = 5 x 10⁻⁶m = 5 μm

The expression for the radius of the Fresnel zones is given by,

rₙ = √[nλd₁d₂/(d₁ + d₂)]

Therefore, the radii of the first five Fresnel zones is,

r₅ = √[5 x 5 x 10⁻⁶x 1 x 1/(1 + 1)]

r₅ = √(25 x 10⁻⁶/2)

r₅ = 3.6 x 10⁻³m

r₅ = 3.6 mm

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Hi there!

Recall that:

Impulse = Δ in momentum = mΔv

Impulse = Force · time

Begin by calculating the change in momentum, or impulse.

I = mΔv = m(vf - vi)

I = (79)(0 - 32) = -2528 Ns

Now, we can use the equation relating force and time to impulse.

I = Ft

Rearrange for time:

I/F = t

-2528/0.25 = -10112 N

**OR, if magnitude ⇒ |-10112| = 10112 N

This question involves the concepts of the equations of motion.

It took him "1.4 s" to stop.

In this scenario, we will use the first equation of motion to find out the time taken by the sled to stop.

\(v_f=v_i+at\)

where,

\(v_f\) = final velocity of the sled = 0 m/s

\(v_i\) = initial velocity of the sled = 282 m/s

a = decceleration = - 201 m/s²

t = time taken = ?

Therefore,

\(0\ m/s=282\ m/s + (-201 m/s^2)t\\\\t=\frac{282\ m/s}{201\ m/s^2}\)

t = 1.4 s

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The vector product of A=2.00i+3.00j+1.00k and B=1.00i-3.00j-2.00k is C=9.00i+4.00j-9.00k.

To find the vector product (also known as the cross product) of two vectors, A and B, we can use the following formula:

C = A × B

Where C is the resultant vector, A and B are the given vectors, and × denotes the cross product.

Given A = 2.00i + 3.00j + 1.00k and B = 1.00i - 3.00j - 2.00k, we can substitute these values into the formula to find the vector product:

C = (2.00i + 3.00j + 1.00k) × (1.00i - 3.00j - 2.00k)

Now, let's expand the cross product using the properties of vector products:

C = (2.00i × 1.00i) + (2.00i × -3.00j) + (2.00i × -2.00k) +

   (3.00j × 1.00i) + (3.00j × -3.00j) + (3.00j × -2.00k) +

   (1.00k × 1.00i) + (1.00k × -3.00j) + (1.00k × -2.00k)

Now, let's calculate each of these cross products:

C = (2.00 × 1.00) \(i^2\) + (2.00 × -3.00) i × j + (2.00 × -2.00) i × k +

   (3.00 × 1.00) j × i + (3.00 × -3.00) \(j^2\) + (3.00 × -2.00) j × k +

   (1.00 × 1.00) k × i + (1.00 × -3.00) k × j + (1.00 × -2.00) \(k^2\)

Since i × j = k, j × k = i, and k × i = j, we can simplify the expression further:

C = 2.00k - 6.00i + 4.00i - 9.00j + k - 3.00j - 2.00j - 2.00k

Combining like terms, we get:

C = (2.00i + 4.00i) + (-6.00i - 9.00j - 3.00j) + (2.00k + k - 2.00k)

Simplifying further:

C = 6.00i - 12.00j + k

Therefore, the vector product of A and B is C = 6.00i - 12.00j + k, which can be written as C = 9.00i + 4.00j - 9.00k in terms of i, j, and k.

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The vector product of A and B is -3i - 5j - 9k.

The vector product, also known as the cross product, of two vectors A and B is denoted as A x B. It is a vector that is perpendicular to both A and B. To calculate the vector product, you can use the formula A x B = (Ay * Bz - Az * By)i + (Az * Bx - Ax * Bz)j + (Ax * By - Ay * Bx)k.

In this case, we have A = 2.00i + 3.00j + 1.00k and B = 1.00i - 3.00j - 2.00k. Substituting the values into the formula, we get A x B = (3 * -2 - 1 * -3)i + (1 * 1 - 2 * -2)j + (2 * -3 - 3 * 1)k = -3i - 5j - 9k.

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Answer:

A car tire is 64.0cm in diameter.

Explanation:

The car tire is traveling at a speed of 16.0m/s .

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =        mass    

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

If A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, then the mass of the air in the classroom is 2322Kg.

Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water).

Given,

Height = 3 m

Width = 20 m

length= 30 m

Density of air = 1.29kg/m³

The  volume of the room = 3×20×30 m³

Volume V = 1800m³

By formula,

Density = Mass/Volume

1.29kg/m³ = Mass/1800m³

Mass of the air = 1.29×1800 = 2322 Kg

The mass of the air is classroom is 2322Kg.

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Answer:

What Is Moral Subjectivism? Moral subjectivism is based on an individual person's perspective of what is right or wrong. An individual can decide for themselves that they approve or disapprove of a certain behavior, and that is what determines if the behavior is right or wrong.

Answer:

I don't know hhaha ammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm I wish I could help

Answer:

Study

Explanation:

From the given analogy, we understand that.

Teachers educates

If this is true,

The students are expected to learn from what the teacher teaches.

This can also be said that, the students studies what the teachers teach

Hence, to complete the analogy.

We have;

(1) Students, Study.

We can also make use of

(2) Students, learn.

(1) & (2) answer the question completely.

Answer:

STUDENT:STUDY OR LEARN

Explanation:

3.38 kg of the unknown substance would contain 845 g of water

Assuming that the mass percentage of water in the unknown substance remains the same, we can use dimensional analysis to determine the mass of the unknown that contains 845 g of water.

Let's call the mass of the unknown substance "m". From the mass percentage calculation, we know that 25% of the unknown's mass is water. Therefore, 75% of the unknown's mass is the substance itself.

We can set up the following proportion:

0.25m = 845 g water

0.75m = ? (mass of the unknown substance)

To solve for the unknown mass, we can rearrange the proportion:

0.75m = (0.75/0.25) * 845 g

0.75m = 2535 g

m = 3380 g

Therefore, 3.38 kg of the unknown substance would contain 845 g of water, assuming the mass percentage of water remains constant.

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